3.63 Switch II

★★

This problem will give you a chance to reverse engineer a switch statement from disassembled machine code. In the following procedure, the body of the switch statement has been omitted:

long switch_prob(long x, long n) {
    long result = x;
    switch(n) {
            /* Fill in code here */
    }
    return result;
}

Below shows the disassembled machine code for the procedure.

The jump table resides in a different area of memory. We can see from the indirect jump on line 5 that the jump table begins at address 0x4006f8. Using the GDB debugger, we can examine the six 8-byte words of memory comprising the jump table with the command x/6gx 0x4006f8. GDB prints the following:

(gdb) x/6gx 0x4006f8
0x4006f8: 0x00000000004005a1 0x00000000004005c3
0x400708: 0x00000000004005a1 0x00000000004005aa
0x400718: 0x00000000004005b2 0x00000000004005bf

Fill in the body of the switch statement with C code that have the same behavior as the machine code.

0000000000400590 <switch_prob>:
    400590: 48 83 ee 3c             sub $0x3c,%rsi
    400594: 48 83 fe 05             cmp $0x5,%rsi
    400598: 77 29                     ja 4005c3 <switch_prob+0x33>
    40059a: ff 24 f5 f8 06 40 00     jmpq *0x4006f8(,%rsi,8)
    4005a1: 48 8d 04 fd 00 00 00     lea 0x0(,%rdi,8),%rax
    4005a8: 00
    4005a9: c3                         retq
    4005aa: 48 89 f8                 mov %rdi,%rax
    4005ad: 48 c1 f8 03             sar $0x3,%rax
    4005b1: c3                         retq
    4005b2: 48 89 f8                 mov %rdi,%rax
    4005b5: 48 c1 e0 04             shl $0x4,%rax
    4005b9: 48 29 f8                 sub %rdi,%rax
    4005bc: 48 89 c7                 mov %rax,%rdi
    4005bf: 48 0f af ff             imul %rdi,%rdi
    4005c3: 48 8d 47 4b             lea 0x4b(%rdi),%rax
    4005c7: c3                         retq

Code:

long switch_prob(long x, long n) {
    long result = x;
    switch(n) {
        case 60:
        case 62:
            result = 8 * x;
            break;
        case 63:
            result = x >> 3;
            break;
        case 64:
            x = 15 * x;
        case 65:
            x *= x;
        default:
            result = 75 + x;
    }
    return result;
}